In Fig. ABC is a triangle in which =2 . D is a point on side BC such that AD bisects and AB=CD. Be is the bisector of . The measure of is:

In Fig. ABC is a triangle in which =2 . D is a point on side BC s…

MCQ

In Fig. $\text{ABC}$ is a triangle in which $\angle\text{B}=2\angle\text{C}. D$ is a point on side $BC$ such that AD bisects $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ Be is the bisector of $\angle\text{B}.$ The measure of $\angle\text{BAC}$ is:

✓
$72^\circ$
B
$95^\circ$
C
$73^\circ$
D
$74^\circ$

✓

Answer

Correct option: A.

$72^\circ$

$\angle\text{ABE}=\angle\text{EBC}$
$(\text{EBC}$ is bisector of $\angle\text{B})$
and $\angle\text{C}=\frac{\angle\text{B}}{2}$
$\Rightarrow\angle\text{EBC}=\angle\text{ECB}$
So $\triangle\text{EBC}$ is isosceles triangle.
$\Rightarrow\text{EB}=\text{EC}\ ....(1)$
Now Consider $\triangle\text{ABE}$ and $\triangle\text{DCE}$
$\text{AB}=\text{DC}\ ($Given$)$
$\text{BE}=\text{CE}\ [$From $(1)]$
$\angle\text{ABE}=\angle\text{DCE} \ ($From above data$)$
So $\triangle\text{ABE}\cong\triangle\text{DCE}$ by $\text{SAS}$ property
$\Rightarrow\text{AE}=\text{DE}$
$\angle\text{BAE}=\angle\text{CDE}=\angle\text{A}$
Now consider $\triangle\text{AED},$
$\text{AE}=\text{DE} \ ($above proved$)$
$\Rightarrow\triangle\text{AED}$ is isosceles triangle
$\Rightarrow\angle\text{EAD}=\angle\text{EDA}=\frac{\angle\text{A}}{2}$
$(AD$ is Bisector of $\angle\text{A})\ ....(2)$
Now, consider $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{C}+\angle\text{C}=180^\circ(\angle\text{B}=2\angle\text{C)}$
$\Rightarrow\angle\text{A}+3\angle\text{C}=180^\circ\ .....(3)$
Consider $\triangle\text{ADE},$
$\Rightarrow\frac{\angle\text{A}}{2}+\angle\text{ADC}+\angle\text{}\text{C}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+(\angle\text{EDA}+\angle\text{CDE})+\angle\text{C}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+\frac{\angle\text{A}}{2}+\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{}A+\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow2\angle\text{A}+\angle\text{C}=180^\circ\ .....(4)$
Right hand side of equations $(3)$ and $(4)$ are equal, hence Left hand side.
$\Rightarrow\angle\text{A}+3\angle\text{C}=2\angle\text{A}+\angle\text{C}$
$\Rightarrow\angle\text{A}=2\angle\text{C}$
Substituting in equation $(3),$
$2\angle\text{C}+3\angle\text{C}=180^\circ$
$\Rightarrow5\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=36^\circ$
$\Rightarrow\angle\text{A}=2\times36^\circ=72^\circ$
Hence, correct option is $(a)$.